Calculate `DeltaG^(@)` for the reaction : `Cu^(2+)(aq) +Fe(s) hArr Fe^(2+)(aq) +Cu(s)`. Given that `E^(@)Cu^(2+)//Cu = 0.34 V`,
`E_(Fe^(+2)//Fe)^(@) =- 0.44 V`

Options

A. 11.44 kJ
B. 180.8 kJ
C. 150.5 kJ
D. 28.5 kJ

Explanation

To calculate `ΔG^(@)` for the reaction `Cu^(2+)(aq) + Fe(s) ⇌ Fe^(2+)(aq) + Cu(s)`, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials
The given half-reactions and their standard electrode potentials are:
1. Reduction of copper:
\[ Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \quad E^(@) = 0.34 \, V \]
2. Oxidation of iron:
\[ Fe(s) \rightarrow Fe^{2+}(aq) + 2e^- \quad E^(@) = -(-0.44 \, V) = 0.44 \, V \]
(Note: The oxidation potential is the negative of the reduction potential.)

### Step 2: Write the overall cell reaction
The overall reaction can be obtained by adding the two half-reactions:
\[ Cu^{2+}(aq) + Fe(s) \rightarrow Fe^{2+}(aq) + Cu(s) \]

### Step 3: Calculate the standard cell potential (E^(@))
The standard cell potential for the overall reaction is calculated as follows:
\[ E^(@)_{cell} = E^(@)_{cathode} - E^(@)_{anode} \]
Here, the cathode is where reduction occurs (copper) and the anode is where oxidation occurs (iron):
\[ E^(@)_{cell} = 0.34 \, V - (-0.44 \, V) = 0.34 \, V + 0.44 \, V = 0.78 \, V \]

### Step 4: Use the Nernst equation to calculate ΔG^(@)
The relationship between Gibbs free energy change and cell potential is given by:
\[ ΔG^(@) = -nFE^(@)_{cell} \]
Where:
- \( n \) = number of moles of electrons transferred (2 for this reaction)
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E^(@)_{cell} \) = standard cell potential calculated in Step 3 (0.78 V)

Substituting the values:
\[ ΔG^(@) = -2 \times 96500 \, C/mol \times 0.78 \, V \]
\[ ΔG^(@) = -2 \times 96500 \times 0.78 = -150.5 \, kJ/mol \]

### Final Answer
Thus, the value of `ΔG^(@)` for the reaction is:
\[ ΔG^(@) = -150.5 \, kJ/mol \]

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